A Primer on Ordinary Differential Equations

Source URL: http://numericalmethods.eng.usf.edu/ Saylor URL: http://www.saylor.org/courses/me205/

Attributed to: University of South Florida: Holistic Numerical Methods Institute Saylor.org

Page 1 of 39

Primer for Ordinary Differential Equations Autar Kaw

After reading this chapter, you should be able to:

1. define an ordinary differential equation, 2. differentiate between an ordinary and partial differential equation, and 3. solve linear ordinary differential equations with fixed constants by using classical

solution and Laplace transform techniques.

Introduction

An equation that consists of derivatives is called a differential equation. Differential

equations have applications in all areas of science and engineering. Mathematical

formulation of most of the physical and engineering problems leads to differential

equations. So, it is important for engineers and scientists to know how to set up

differential equations and solve them.

Differential equations are of two types

(A) ordinary differential equations (ODE) (B) partial differential equations (PDE)

An ordinary differential equation is that in which all the derivatives are with respect to a

single independent variable. Examples of ordinary differential equations include

022

2

=++ ydx

dy

dx

yd, 4)0( ,2)0( == ydx

dy,

,sin532

2

3

3

xydx

dy

dx

yd

dx

yd=+++ ,12)0(

2

2

=dx

yd 2)0( =dx

dy, 4)0( =y

Ordinary differential equations are classified in terms of order and degree. Order of an

ordinary differential equation is the same as the highest derivative and the degree of an

ordinary differential equation is the power of highest derivative.



Page 2 of 39

Thus the differential equation,

xexydx

dyx

dx

ydx

dx

ydx =+++

2

22

3

33

is of order 3 and degree 1, whereas the differential equation

xdx

dyx

dx

dysin1 2

2

=+

+

is of order 1 and degree 2.

An engineer’s approach to differential equations is different from a mathematician.

While, the latter is interested in the mathematical solution, an engineer should be able

to interpret the result physically. So, an engineer’s approach can be divided into three

phases:

a) formulation of a differential equation from a given physical situation, b) solving the differential equation and evaluating the constants, using given

conditions, and c) interpreting the results physically for implementation.

Formulation of differential equations

As discussed above, the formulation of a differential equation is based on a given

physical situation. This can be illustrated by a spring-mass-damper system.



Page 3 of 39

Above is the schematic diagram of a spring-mass-damper system. A block is

suspended freely using a spring. As most physical systems involve some kind of

damping - viscous damping, dry damping, magnetic damping, etc., a damper or dashpot

is attached to account for viscous damping.

Let the mass of the block be M , the spring constant be K , and the damper

coefficient be b . If we measure displacement from the static equilibrium position we

need not consider gravitational force as it is balanced by tension in the spring at

equilibrium.

Below is the free body diagram of the block at static and dynamic equilibrium.

So, the equation of motion is given by

DS FFMa += (1)

where

SF is the restoring force due to spring.

DF is the damping force due to the damper.

a is the acceleration.

The restoring force in the spring is given by

K b

x

Figure 1 Spring-mass damper system.

M



Page 4 of 39

KxFS −= (2)

as the restoring force is proportional to displacement and it is negative as it opposes the

motion. The damping force in the damper is given by

bvFD −= (3)

as the damping force is directly proportional to velocity and also opposes motion.

Therefore, the equation of motion can be written as

bvKxMa −−=

(4)

Since

2

2

dt

xda = and

dt

dxv =

from Equation (4), we get

SF

Ma

DF

T

Mg

Figure 2 Free body diagram of spring-mass-damper system.

Static Dynamic



Page 5 of 39

dt

dxbKx

dt

xdM −−=

2

2

02

2

=++ Kxdt

dxb

dt

xdM (5)

This is an ordinary differential equation of second order and of degree one.

Solution to linear ordinary differential equations

In this section we discuss two techniques used to solve ordinary differential equations

(A) Classical technique (B) Laplace transform technique

Classical Technique

The general form of a linear ordinary differential equation with constant coefficients is

given by

)(......... 122

2

31

1

xFykdx

dyk

dx

ydk

dx

ydk

dx

ydn

n

nn

n

=+++++−

−

(6)

The general solution contains two parts

PH yyy += (7)

where

Hy is the homogeneous part of the solution and

Py is the particular part of the solution.

The homogeneous part of the solution Hy is that part of the solution that gives zero

when substituted in the left hand side of the equation. So, Hy is solution of the equation

0......... 122

2

31

1

=+++++−

−

ykdx

dyk

dx

ydk

dx

ydk

dx

ydn

n

nn

n

(8)



Page 6 of 39

The above equation can be symbolically written as

0................. 12

1 =++++ − ykDykyDkyD n

n

n (9)

0).................( 12

1 =++++ − ykDkDkD n

n

n (10)

where,

n

nn

dx

dD =

(11)

.

.

.

1

11

−

−− =

n

nn

dx

dD

operating on y is the same as

),( 1rD − )( 2rD − , )( nrD −

operating one after the other in any order, where

)(....,),........(),( 21 nrDrDrD −−−

are factors of

0............... 12

1 =++++ − kDkDkD n

n

n (12)

To illustrate

0)23( 2 =+− yDD

is same as

0)1)(2( =−− yDD

0)2)(1( =−− yDD



Page 7 of 39

Therefore,

0)....................( 12

1 =++++ − ykDkDkD n

n

n (13)

is same as

0).........().........)(( 11 =−−− − yrDrDrD nn (14)

operating one after the other in any order.

Case 1: Roots are real and distinct

The entire left hand side becomes zero if ( ) 01 =− yrD . Therefore, the solution to

( ) 01 =− yrD is a solution to a homogeneous equation. ( ) 01 =− yrD is called Leibnitz’s

linear differential equation of first order and its solution is

( ) 01 =− yrD (15)

yrdx

dy1= (16)

dxry

dy1= (17)

Integrating both sides we get

cxry += 1ln (18)

xrcey 1= (19)

Since any of the n factors can be placed before y , there are n different solutions

corresponding to n different factors given by

xrxrxr

n

xr

n eCeCeCeC nn 121

121 ,.....,,........., −−

where

121, ,..,,......... rrrr nn − are the roots of Equation (12) and

121, ,,......, CCCC nn − are constants.



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We get the general solution for a homogeneous equation by superimposing the

individual Leibnitz’s solutions. Therefore

xr

n

xr

n

xrxr

Hnn eCeCeCeCy ++++= −

−121

121 ............. (20)

Case 2: Roots are real and identical

If two roots of a homogeneous equation are equal, say 21 rr = , then

0))(...(..........).........)(( 111 =−−−− − yrDrDrDrD nn (21)

Let’s work at

0))(( 11 =−− yrDrD (22)

If

zyrD =− )( 1 (23)

then

0)( 1 =− zrD

xreCz 1

2= (24)

Now substituting the solution from Equation (24) in Equation (23)

xreCyrD 1

21)( =−

xreCyr

dx

dy1

21 =−

2111 Cyer

dx

dye

xrxr =− −−

2

)( 1

Cdx

yedxr

=−

dxCyed xr

2)( 1 =− (25)



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Integrating both sides of Equation (25), we get

121 CxCye xr +=−

xreCxCy 1)( 12 += (26)

Therefore the final homogeneous solution is given by

( ) xr

n

xrxr

HneCeCexCCy ++++= ...31

321

(27)

Similarly, if m roots are equal the solution is given by

( ) xr

n

xr

m

xrm

mHnmm eCeCexCxCxCCy +++++++= +

+− .......... 1

1

12

321 (28)

Case 3: Roots are complex

If one pair of roots is complex, say βα ir +=1 and βα ir −=2 ,

where

1−=i

then

( ) ( ) xr

n

xrxixi

HneCeCeCeCy ++++= −+ ......3

321

βαβα (29)

Since

xixe xi βββ sincos += , and (30a)

xixe xi βββ sincos −=− (30b)

then

( ) ( ) xr

n

xrxx

HneCeCxixeCxixeCy +++−++= .........sincossincos 3

321 ββββ αα

( ) ( ) xr

n

xrxx neCeCxeCCixeCC +++−++= .........sincos 3

32121 ββ αα

( ) xr

n

xrx neCeCxBxAe ++++= ........sincos 3

3ββα (31)



Page 10 of 39

where

21 CCA += and

)( 21 CCiB −= (32)

Now, let us look at how the particular part of the solution is found. Consider the general

form of the ordinary differential equation

( ) XykDkDkD n

n

n

n

n =++++ −−

−1

2

1

1 .......... (33)

The particular part of the solution Py is that part of solution that gives X when

substituted for y in the above equation, that is,

( ) XykDkDkD P

n

n

n

n

n =++++ −−

−1

2

1

1 ...... (34)

Sample Case 1

When axeX = , the particular part of the solution is of the form axAe . We can find A by

substituting axAey = in the left hand side of the differential equation and equating

coefficients.

Example 1

Solve

xeydx

dy −=+ 23 , 5)0( =y

Solution

The homogeneous solution for the above equation is given by

( ) 023 =+ yD

The characteristic equation for the above equation is given by

023 =+r

The solution to the equation is



Page 11 of 39

666667.0−=r

x

H Cey 666667.0−=

The particular part of the solution is of the form xAe−

( ) xx

x

eAedx

Aed −−−

=+ 23

xxx eAeAe −−− =+− 23

xx eAe −− =−

1−=A

Hence the particular part of the solution is

x

P ey −−=

The complete solution is given by

PH yyy +=

xx eCe −− −= 666667.0

The constant C can be obtained by using the initial condition 5)0( =y

( ) 50 00666667.0 =−= −×− eCey

51 =−C

6=C

The complete solution is

xx eey −− −= 666667.06

Example 2

Solve



Page 12 of 39

xeydx

dy 5.132 −=+ , 5)0( =y

Solution

The homogeneous solution for the above equation is given by

( ) 032 =+ yD

The characteristic equation for the above equation is given by

032 =+r

The solution to the equation is

5.1−=r

x

H Cey 5.1−=

Based on the forcing function of the ordinary differential equations, the particular part of

the solution is of the form xAe 5.1− , but since that is part of the form of the homogeneous

part of the solution, we need to choose the next independent solution, that is,

x

P Axey 5.1−=

To find A , we substitute this solution in the ordinary differential equation as

( ) xx

x

eAxedx

Axed 5.15.15.1

32 −−−

=+

xxxx eAxeAxeAe 5.15.15.15.1 332 −−−− =+−

xx eAe 5.15.12 −− =

5.0=A


x

P xey 5.15.0 −=


PH yyy +=



Page 13 of 39

xx xeCe 5.15.1 5.0 −− +=

The constant C is obtained by using the initial condition 5)0( =y .

( ) 5)0(5.00 )0(5.1)0(5.1 =+= −− eCey

50 =+C

5=C


xx xeey 5.15.1 5.05 −− +=

Sample Case 2 When

)sin(axX = or )cos(ax ,

the particular part of the solution is of the form

)cos()sin( axBaxA + .

We can get Aand B by substituting )cos()sin( axBaxAy += in the left hand side of the

differential equation and equating coefficients.

Example 3

Solve

xydx

dy

dx

ydsin125.332

2

2

=++ , 3)0( ,5)0( === xdx

dyy

Solution

The homogeneous equation is given by

0)125.332( 2 =++ yDD

The characteristic equation is



Page 14 of 39

0125.332 2 =++ rr

The roots of the characteristic equation are

22

125.32433 2

×××−±−

=r

4

2593

−±−=

4

163

−±−=

4

43

i±−=

i±−= 75.0

Therefore the homogeneous part of the solution is given by

)sincos( 21

75.0 xKxKey x

H += −

The particular part of the solution is of the form

xBxAyP cossin +=

( ) ( ) xxBxAxBxAdx

dxBxA

dx

dsin)cossin(125.3cossin3cossin2

2

2

=+++++

( ) xxBxAxBxAxBxAdx

dsin)cossin(125.3)sincos(3sincos2 =++−+−

xxBxAxBxAxBxA sin)cossin(125.3)sincos(3)cossin(2 =++−+−−

xxABxBA sincos)3125.1(sin)3125.1( =++−

Equating coefficients of xsin and xcos on both sides, we get

13125.1 =− BA

03125.1 =+ AB

Solving the above two simultaneous linear equations we get



Page 15 of 39

109589.0=A

292237.0−=B

Hence

xxyP cos292237.0sin109589.0 −=


)cos292237.0sin109589.0()sincos( 21

75.0 xxxKxKey x −++= −

To find 1K and 2K we use the initial conditions

3)0( ,5)0( === xdx

dyy

From 5)0( =y we get

))0cos(292237.0)0sin(109589.0())0sin()0cos((5 21

)0(75.0 −++= − KKe

292237.05 1 −= K

292237.51 =K

xx

xKxKexKxKedx

dy xx

sin292237.0cos109589.0

)cossin()sincos(75.0 21

75.0

21

75.0

++

+−++−= −−

From

,3)0( ==xdx

dy

we get

)0sin(292237.0)0cos(109589.0

))0cos()0sin(( ))0sin()0cos((75.03 21

)0(75.0

21

)0(75.0

++

+−++−= −− KKeKKe

109589.075.03 21 ++−= KK

109589.0)292237.5(75.03 2 ++−= K



Page 16 of 39

859588.62 =K


xxxxey x cos292237.0sin109589.0)sin859588.6cos292237.5(75.0 −++= −

Example 4

Solve

)cos(125.3622

2

xydx

dy

dx

yd=++ , 3)0( ,5)0( === x

dx

dyy

Solution

The homogeneous part of the equations is given by

0)125.362( 2 =++ yDD

The characteristic equation is given by

0125.362 2 =++ rr

)2(2

)125.3)(2(466 2 −±−=r

4

25366 −±−=

4

116 ±−=

829156.05.1 ±−=

329156.2,670844.0 −−=

Therefore, the homogeneous solution Hy is given by

xx

H eKeKy 329156.2

2

670845.0

1

−− +=




Page 17 of 39

xBxAyP cossin +=

Substituting the particular part of the solution in the differential equation,

xxBxA

xBxAdx

dxBxA

dx

d

cos)cossin(125.3

)cossin(6)cossin(22

2

=++

+++

xxBxA

xBxAxBxAdx

d

cos)cossin(125.3

)sincos(6)sincos(2

=++

−+−

xxBxA

xBxAxBxA

cos)cossin(125.3

)sincos(6)cossin(2

=++

−+−−

xxABxBA coscos)6125.1(sin)6125.1( =++−

Equating coefficients of xcos and xsin we get

06125.1

16125.1

=−

=+

BA

AB

The solution to the above two simultaneous linear equations are

0301887.0

161006.0

=

=

B

A


xxyP cos0301887.0sin161006.0 +=

Therefore the complete solution is

PH yyy +=

xxeKeKy xx cos0301887.0sin161006.0)( 329156.2

2

670845.0

1 +++= −−

Constants 1K and 2K can be determined using initial conditions. From 5)0( =y ,

50301887.0)0( 21 =++= KKy

969811.40301887.0521 =−=+ KK



Page 18 of 39

Now

xx

eKeKdx

dy xx

sin0301887.0cos161006.0

329156.2670845.0 )329156.2(

2

)670845.0(

1

−+

−−= −−

From 3)0( ==xdx

dy

3161006.0329156.2670845.0 21 =+−− KK

161006.03329156.2670845.0 21 +−=+ KK

838994.2329156.2670845.0 21 −=+ KK

We have two linear equations with two unknowns

969811.421 =+ KK

838994.2329156.2670845.0 21 −=+ KK

Solving the above two simultaneous linear equations, we get

692253.81 =K

722442.32 −=K


.cos0301887.0sin161006.0

)722442.3692253.8( 329156.2670845.0

xx

eey xx

++

−= −−

Sample Case 3 When

bxeX ax sin= or bxeax cos ,

the particular part of the solution is of the form

)cossin( bxBbxAeax + ,



Page 19 of 39

we can get Aand B by substituting

)cossin( bxBbxAey ax +=

in the left hand side of differential equation and equating coefficients.

Example 5

Solve

xeydx

dy

dx

yd x sin125.3522

2−=++ , 3)0( ,5)0( === x

dx

dyy

Solution

The homogeneous equation is given by

0)125.352( 2 =++ yDD

The characteristic equation is given by

0125.352 2 =++ rr

)2(2

)125.3)(2(455 2 −±−=r

4

25255

−±−=

4

05 ±−=

25.1,25.1 −−=

Since roots are repeated, the homogeneous solution Hy is given by

x

H exKKy )25.1(

21 )( −+=


)cossin( xBxAey x

P += −



Page 20 of 39

Substituting the particular part of the solution in the ordinary differential equation

xexBxAe

xBxAedx

dxBxAe

dx

d

xx

xx

sin)}cossin({125.3

)}cossin({5)}cossin({22

2

−−

−−

=++

+++

xexBxAexBxAexBxAe

xBxAexBxAedx

d

xxxx

xx

sin)cossin(125.3)}sincos()cossin({5

)}sincos()cossin({2

−−−−

−−

=++−++−+

−++−

xexBxAexBxAexBxAe

xBxAexBxAexBxAexBxAe

xxxx

xxxx

sin)cossin(125.3)}sincos()cossin({5

)}cossin()sincos()sincos()cossin({2

−−−−

−−−−

=++−++−+

+−−−−−+

xexBxAexBxAexxxsin)sincos()cossin(875.1

−−− =−++−

xxBxAxBxA sin)sincos()cossin(875.1 =−++−

xxBAxBA sincos)875.1(sin)875.1( =−++−

Equating coefficients of xcos and xsin on both sides we get

0875.1 =− BA

1875.1 −=+ BA

Solving the above two simultaneous linear equations we get

415224.0−=A and

221453.0−=B

Hence,

)cos221453.0sin415224.0( xxey x

P +−= −

Therefore complete solution is given by

PH yyy +=

)cos221453.0sin415224.0()( 25.1

21 xxeexKKy xx +−+= −−

Constants 1K and 2K can be determined using initial conditions,

From ,5)0( =y we get



Page 21 of 39

5221453.01 =−K

221453.51 =K

Now

)cos221453.0sin415224.0()sin221453.0cos415224.0(

25.125.1 25.1

2

25.1

2

25.1

1

xxexxe

eKxeKeKdx

dy

xx

xxx

++−

−+−−=

−−

−−−

From ,3)0( =dx

dy we get

3)0cos(221453.0)0sin(415224.0())0sin(221453.0)0cos(415224.0(

)0(25.125.1

00

)0(25.1

2

)0(25.1

2

)0(25.1

1

=++−−

+−− −−−

ee

eKeKeK

3415224.0221453.025.1 21 =−++− KK

193771.325.1 21 =+− KK

193771.3)221453.5(25.1 2 =+− K

720582.92 =K

Substituting

221453.51 =K and

720582.92 =K

in the solution, we get

)cos221453.0sin415224.0()720582.9221453.5( 25.1 xxeexy xx +−+= −−

The forms of the particular part of the solution for different right hand sides of ordinary

differential equations are given below

X ( )xyP



Page 22 of 39

2

210 xaxaa ++ 2

210 xbxbb ++

axe axAe

)sin(bx )cos()sin( bxBbxA +

)sin(bxeax ( ))cos()sin( bxBbxAeax +

)cos(bx )cos()sin( bxBbxA +

)cos(bxeax ( ))cos()sin( bxBbxAeax +

Laplace Transforms

If )(xfy = is defined at all positive values of x , the Laplace transform denoted by )(sY

is given by

dxxfexfLsY sx )()}({)(0

∫∞

−== (35)

where s is a parameter, which can be a real or complex number. We can get back

)(xf by taking the inverse Laplace transform of )(sY .

)()}({1 xfsYL =− (36)

Laplace transforms are very useful in solving differential equations. They give the

solution directly without the necessity of evaluating arbitrary constants separately.

The following are Laplace transforms of some elementary functions

sL

1)1( =

1

!)(

+=

n

n

s

nxL , where ....3,2,1,0=n

as

eL ax

−=

1)(



Page 23 of 39

22

)(sinas

aaxL

+=

22

)(cosas

saxL

+=

22

)(sinhas

aaxL

−=

22

)(coshas

saxL

−= (37)

The following are the inverse Laplace transforms of some common functions

111 =

−

sL

axeas

L =

−

− 11

( )!1

1 11

−=

−−

n

x

sL

n

n, where ......3,2,1=n

( ) ( )!11 1

1

−=

−

−−

n

xe

asL

nax

n

axaas

L sin11

22

1 =

+

−

axas

sL cos

22

1 =

+

−

axaas

L sinh11

22

1 =

−

−

atas

sL cosh

22

1 =

−

−



Page 24 of 39

( )

bxebbas

L ax sin11

22

1 =

+−−

( )

bxebas

asL ax cos

22

1 =

+−

−−

( )

axxaas

sL sin

2

1222

1 =

+

−

(38)

Properties of Laplace transforms

Linear property

If cba , , are constants and ),( ),( xgxf and )(xh are functions of x then

))(())(())(()]()()([ xhcLxgbLxfaLxchxbgxafL ++=++ (39)

Shifting property If

)()}({ sYxfL = (40)

then

)()}({ asYxfeL at −= (41)

Using shifting property we get

( )( ) 1

!+−

=n

nax

as

nxeL , 0≥n

( )( ) 22

sinbas

bbxeL ax

+−=

( )( ) 22

cosbas

asbxeL ax

+−

−=



Page 25 of 39

( )( ) 22

sinhbas

bbxeL ax

−−=

( )( ) 22

coshbas

asbxeL ax

−−

−= (42)

Scaling property If

)()}({ sYxfL = (43)

then

=a

sY

aaxfL

1)}({ (44)

Laplace transforms of derivatives

If the first n derivatives of )(xf are continuous then

∫∞

−=0

)()}({ dxxfexfL nsxn (45)

Using integration by parts we get

∫

∫∞

−

∞∞

−−−−−

−−−−−

−−+

−−++−+

−−=

0

001132

21

)()()1(

)()()1(......)()(

)()()()(

dxxfes

xfesxfes

xfesxfedxxfe

sxnn

sxnnnsx

nsxnsx

nsx

∫∞

−−−−− +−−−−−=0

13221 )()0(.............)0()0()0( dxxfesfsfssff sxnnnnn

)0(........)0()0()0()( 13221 fsfssffsYs nnnnn −−−− −−−−−=

(46)



Page 26 of 39

Laplace transform technique to solve ordinary differential equations

The following are steps to solve ordinary differential equations using the Laplace

transform method

(A) Take the Laplace transform of both sides of ordinary differential equations. (B) Express )(sY as a function of s .

(C) Take the inverse Laplace transform on both sides to get the solution. Let us solve Examples 1 through 4 using the Laplace transform method.

Example 6

Solve

xeydx

dy −=+ 23 , 5)0( =y

Solution

Taking the Laplace transform of both sides, we get

( )xeLydx

dyL −=

+ 23

1

1)(2)]0()([3

+=+−s

sYyssY

Using the initial condition, 5)0( =y we get

1

1)(2]5)([3

+=+−s

sYssY

151

1)()23( +

+=+s

sYs

1

1615)()23(

++

=+s

ssYs

)23)(1(

1615)(

+++

=ss

ssY

Writing the expression for )(sY in terms of partial fractions



Page 27 of 39

231)23)(1(

1615

++

+=

+++

s

B

s

A

ss

s

)23)(1(

23

)23)(1(

1615

+++++

=++

+ss

BBsAAs

ss

s

BBsAAss +++=+ 231615

Equating coefficients of 1s and 0s gives

153 =+ BA

162 =+ BA

The solution to the above two simultaneous linear equations is

1−=A

18=B

23

18

1

1)(

++

+−

=ss

sY

666667.0

6

1

1

++

+−

=ss

Taking the inverse Laplace transform on both sides

+

+

+−

= −−−

666667.0

6

1

1)}({ 111

sL

sLsYL

Since

ateas

L −− =

+11

The solution is given by

xx eexy 666667.06)( −− +−=

Example 7



Page 28 of 39

Solve

xeydx

dy 5.132 −=+ , 5)0( =y

Solution

Taking the Laplace transform of both sides, we get

( )xeLydx

dyL 5.132 −=

+

5.1

1)(3)]0()([2

+=+−s

sYyssY

Using the initial condition 5)0( =y , we get

5.1

1)(3]5)([2

+=+−s

sYssY

105.1

1)()32( +

+=+s

sYs

5.1

1610)()32(

++

=+s

ssYs

)32)(5.1(

1610)(

+++

=ss

ssY

)5.1)(5.1(2

1610

+++

=ss

s

2)5.1(2

1610

+

+=

s

s

2)5.1(

85

+

+=

s

s




Page 29 of 39

22 )5.1(5.1)5.1(

85

++

+=

+

+

s

B

s

A

s

s

22 )5.1(

5.1

)5.1(

85

+

++=

+

+

s

BAAs

s

s

BAAss ++=+ 5.185

Equating coefficients of 1s and 0s gives

5=A

85.1 =+ BA

The solution to the above two simultaneous linear equations is

5=A

5.0=B

2)5.1(

5.0

5.1

5)(

++

+=

sssY


++

+

= −−−2

111

)5.1(

5.0

5.1

5)}({

sL

sLsYL

Since

axeas

L −− =

+11 and axxe

asL −− =

+ 2

1

)(

1

The solution is given by

xx xeexy 5.15.1 5.05)( −− +=



Page 30 of 39

Example 8

Solve

xydx

dy

dx

ydsin125.332

2

2

=++ , 3)0( ,5)0( === xdx

dyy

Solution

Taking the Laplace transform of both sides

( )xLydx

dy

dx

ydL sin125.332

2

2

=

++

and knowing

2

2

dx

ydL ( ) ( ) ( )002 =−−= x

dx

dysysYs

dx

dyL ( ) ( )0yssY −=

1

1)(sin

2 +=s

xL

we get

[ ]1

1)(125.3)0()(3)0()0()(2

2

2

+=+−+

=−−s

sYyssYxdx

dysysYs

[ ] [ ]1

1)(125.35)(335)(2

2

2

+=+−+−−s

sYssYssYs

( )[ ]1

12110)(125.332

2 +=−−++s

ssYss

( )[ ] 21101

1)(125.332

2++

+=++ ss

sYss

[ ])1(

21101022)(125.332

2

232

+

+++=++

s

ssssYss



Page 31 of 39

( )( )125.3321

22102110)(

22

23

+++

+++=

sss

ssssY


( ) ( ) ( )( )125.3321

22102110

1125.332 22

23

22 +++

+++=

+

++

++

+

sss

sss

s

DCs

ss

BAs

( )( )

( )( )125.3321

22102110

1125.332

125.332125.332

22

23

22

22323

+++

+++=

+++

+++++++++

sss

sss

sss

DDsDsCsCsCsBBsAsAs

( ) ( ) ( ) ( )( )( )

( )( )125.3321

22102110

125.3321

125.33125.3232

22

23

22

23

+++

+++=

+++

+++++++++

sss

sss

sss

DBsDCAsDCBsCA

Equating terms of 3s , 12 , ss and 0s gives

102 =+ CA

2123 =++ DCB

103125.3 =++ DCA

22125.3 =+ DB

The solution to the above four simultaneous linear equations is

584474.10=A

657534.21=B

292237.0−=C

109589.0=D

Hence



Page 32 of 39

1

109589.0292237.0

125.332

657534.21584474.10)(

22 +

+−+

++

+=

s

s

ss

ssY

( ) }1)75.0{(2}1)5625.05.1{(2125.332 222 ++=+++=++ sssss

1

109589.0292237.0

}1)75.0{(2

719179.13)75.0(584474.10)(

22 +

+−+

++

++=

s

s

s

ssY

)1(

109589.0

)1(

292237.0

}1)75.0{(

859589.6

}1)75.0{(

)75.0(292237.5

2222 ++

+−

+++

++

+=

ss

s

ss

s

Taking the inverse Laplace transform of both sides

+

+

+

−

+++

++

+=

−−

−−−

1

109589.0

1

292237.0

1)75.0{(

859589.6

}1)75.0{(

)75.0(292237.5)}({

2

1

2

1

2

1

2

11

sL

s

sL

sL

s

sLsYL

+

+

+

−

+++

++

+=

−−

−−−

1

1109589.0

1292237.0

1)75.0{(

1859589.6

}1)75.0{(

75.0292237.5)}({

2

1

2

1

2

1

2

11

sL

s

sL

sL

s

sLsYL

Since

( )

bxebas

asL ax cos

22

1 −− =

++

+

( )

bxebas

bL ax sin

22

1 −− =

++

axas

L sin1

22

1 =

+

−

axas

sL cos

22

1 =

+

−




Page 33 of 39

xx

xexexy xx

sin109589.0cos292237.0

sin8595859.6cos292237.5)( 75.075.0

+−

+= −−

( ) xxxxe x sin109589.0cos292237.0sin859589.6cos292237.5 75.0 +−+= −

Example 9

Solve

xydx

dy

dx

ydcos125.362

2

2

=++ , 3)0( ,5)0( === xdx

dyy

Solution


( )xLydx

dy

dx

ydL cos125.362

2

2

=

++

and knowing

2

2

dx

ydL ( ) ( ) ( )002 =−−= x

dx

dysysYs

dx

dyL ( ) ( )0yssY −=

1

)(cos2 +

=s

sxL

we get

[ ]1

)(125.3)0()(6)0()0()(22

2

+=+−+

=−−s

ssYyssYx

dx

dysysYs

[ ] [ ]1

)(125.35)(635)(22

2

+=+−+−−s

ssYssYssYs

[ ] 3610

1)(125.3)62(

2++

+=++ ss

ssYss



Page 34 of 39

[ ]1

36111036)(125.362

2

232

+

+++=++

s

ssssYss

( )( )125.3621

36113610)(

22

23

+++

+++=

sss

ssssY


( ) ( ) ( )( )125.3621

36113610

1125.362 22

23

22 +++

+++=

+

++

++

+

sss

sss

s

DCs

ss

BAs

( )( )

( )( )125.3621

36113610

1125.362

125.362125.362

22

23

22

22323

+++

+++=

+++

+++++++++

sss

sss

sss

DDsDsCsCsCsBBsAsAs

( ) ( ) ( ) ( )( )( )

( )( )125.3621

36113610

125.3621

125.36125.3262

22

23

22

23

+++

+++=

+++

+++++++++

sss

sss

sss

DBsDCAsDCBsCA

Equating terms of 3s , 12 , ss and 0s gives

102 =+ CA

3626 =++ DCB

116125.3 =++ DCA

36125.3 =+ DB


939622.9=A

496855.35=B

0301886.0=C

161006.0=D



Page 35 of 39

Then

1

161006.00301886.0

125.362

496855.35939622.9)(

22 +

++

++

+=

s

s

ss

ssY

( ) }829156.0)5.1{(2}6875.0)25.23{(2125.362 2222 −+=−++=++ sssss

1

161006.00301886.0

}829156.0)5.1{(2

587422.20)5.1(939622.9)(

222 +

++

−+

++=

s

s

s

ssY

1

161006.0

1

0301886.0

}829156.0)5.1{(

293711.10

}829156.0)5.1{(

)5.1(969811.4

22

2222

++

++

−++

−+

+=

ss

s

ss

s


+

+

+

+

−++

−+

+=

−−

−−−

1

161006.0

1

0301886.0

829156.0)5.1{(

293711.10

}829156.0)5.1{(

)5.1(969811.4)}({

2

1

2

1

22

1

22

11

sL

s

sL

sL

s

sLsYL

−+

+

−++

= −−22

1

22

1

829156.0)5.1(

1293711.10

829156.0)5.1(

)5.1(969811.4

sL

s

sL

++

++ −−

)1(

1161006.0

)1(0301886.0

2

1

2

1

sL

s

sL

Since

( )

bxebas

asL ax cosh

22

1 −− =

−+

+

( )

bxebbas

L ax sinh11

22

1 −− =

−+

axaas

L sin11

22

1 =

+

−



Page 36 of 39

axas

sL cos

22

1 =

+

−


xx

xexexy xx

sin161006.0cos0301886.0

)829156.0sinh(829156.0

293711.10)829156.0cosh(969811.4)( 5.15.1

++

+= −−

xx

eeeee

xxxxx

sin161006.0cos030188.0

2414685.12

2969811.4

829156.0829156.0829156.0829156.05.1

++

−+

+=

−−−

( )

x

xeee xxx

sin161006.0

cos0301886.0722437.3692248.8 829156.0829156.05.1

+

+−= −−

Example 10

Solve

xeydx

dy

dx

yd x sin125.3522

2−=++ , 3)0( ,5)0( === x

dx

dyy

Solution


( )xeLydx

dy

dx

ydL x sin125.352

2

2−=

++

knowing

2

2

dx

ydL ( ) ( ) ( )002 =−−= x

dx

dysysYs

dx

dyL ( ) ( )0yssY −=



Page 37 of 39

1)1(

1)sin(

2 ++=−

sxeL x

we get

[ ]

[ ] [ ]1)1(

1)(125.35)(535)(2

1)1(

1)(125.3)0()(5)0()0()(2

2

2

2

2

++=+−+−−

++=+−+

=−−

ssYssYssYs

ssYyssYx

dx

dysysYs

( )[ ]1)1(

13110)(125.352

2 ++=−−++

sssYss

[ ] 31101)1(

1)(125.3)52(

2++

++=++ s

ssYss

[ ]22

51821063)(125.352

2

232

++

+++=++

ss

ssssYss

( )( )125.35222

63825110)(

22

23

++++

+++=

ssss

ssssY


( )( )125.35222

63825110

22125.352 22

23

22 ++++

+++=

++

++

++

+

ssss

sss

ss

DCs

ss

BAs

( )( )

( )( )125.35222

63825110

22125.352

2222125.352125.352

22

23

22

223223

++++

+++=

++++

+++++++++++

ssss

sss

ssss

BBsBsAsAsAsDDsDsCsCsCs

( ) ( ) ( ) ( )( )( )

( )( )125.35222

63825110

125.35222

2125.3225125.32252

22

23

22

23

++++

+++=

++++

+++++++++++

ssss

sss

ssss

BDsBADCsBADCsAC



Page 38 of 39

Equating terms of 3s , 12 , ss and 0s gives four simultaneous linear equations

102 =+ AC

51225 =+++ BADC

82225125.3 =+++ BADC

632125.3 =+ BD


442906.10=A

494809.32=B

221453.0−=C

636678.0−=D

Then

22

636678.0221453.0

125.352

494809.32442906.10)(

22 ++

−−+

++

+=

ss

s

ss

ssY

( ) 222 )25.1(2)}5625.15.2{(2125.352 +=++=++ sssss

1)1(

415225.0)1(221453.0

)25.1(2

441176.19)25.1(442906.10)(

22 ++

−+−+

+

++=

s

s

s

ssY

1)1(

415225.0

1)1(

)1(221453.0

)25.1(

720588.9

)25.1(

)25.1(221453.5

2222 ++−

++

+−

++

+

+=

ss

s

ss

s


++−

+++

−

++

+=

−−

−−−

1)1(

415225.0

1)1(

)1(221453.0

)25.1(

720588.9

)25.1(

221453.5)}({

2

1

2

1

2

111

sL

s

sL

sL

sLsYL



Page 39 of 39

++−

+++

−

++

+=

−−

−−

1)1(

1415225.0

1)1(

)1(221453.0

)25.1(

1720588.9

)25.1(

1221453.5

2

1

2

1

2

11

sL

s

sL

sL

sL

Since

( )

bxebas

asL ax cos

22

1 −− =

++

+

( )

bxebas

bL ax sin

22

1 −− =

++

axeas

L −− =

+11

)!1()(

1 11

−=

+

−−−

n

xe

asL

nax

n


xe

xexeexy

x

xxx

sin415225.0

cos221453.0720588.9221453.5)( 25.125.1

−

−−−

−

−+=

( ) )sin415225.0cos221453.0(720588.9221453.5 25.1 xxexe xx −−++= −

Documents

A Primer on Ordinary Differential Equations